Trouble with deriving the Euler-Poisson equation using Euler's approach

Question

"A First Course in the Calculus of Variations" by Mark Kot, §2.2. Euler’s approach describes a method for deriving the Euler-Lagrange equation which I shall call "geometric". It is also described here and in this post. I would like to derive the general version of this equation, also called the Euler-Poisson equation

$$\sum_{k=0}^{n}\left( -1 \right)^{k}\frac{d^{k}}{dx^{k}}\left( \frac{\partial F}{\partial y^{(k)}} \right)=0$$

for my students, they didn't like Lagrange's approach.

I seem to be having issues with going above the first derivative, $y'$, though. How does one derive the Euler-Poisson equation using the geometric approach?

Answer 1

If I understand correctly, you want to derive the EL equations in the discrete setting. You just need to replace the function $y$ by a sequence. You can then write your action as: $$ S = \sum_{i\in\mathbb Z} G(y_i,...,y_{i+n}) $$ Varying $y$ gives: $$ \frac{\partial S}{\partial y_i} = \sum_{k=0}^n\partial_k G(y_{i-k},...,y_{i-k+n}) $$ To make the link with the continuous limit, just rewrite: $$ G(y_i,...y_{i+n}) = F(y_i^{(0)},...,y_i^{(n)})\Delta x $$ with $y_i^{(k)}$ the $k$-th discrete derivative at $i$: $$ y_i^{(k)} = \frac{1}{\Delta x^k}\sum_{l=0}^k a_{kl}y_{i+l} $$ with the $a$ are the forward finite difference coefficients: $$ \begin{align} k&=0 & 1 \\ k&=1 & -1 && 1\\ k&=2 & 1 && -2 && 1 \\ \end{align} \\ a_{kl} = (-1)^{k-l}\binom{k}{l} $$ The variation is now: $$ \begin{align} \left(\frac{\delta S}{\delta y}\right)_i &= \frac{1}{\Delta x}\frac{\partial S}{\partial y_i} \\ &= \frac{1}{\Delta x}\sum_{k=0}^n\sum_{l=k}^n\frac{\partial F}{\partial y^{(l)}}(y_{i-k}^{(0)},...,y_{i-k}^{(n)})\frac{\partial y_{i-k}^{(l)}}{\partial y_i} \Delta x \\ &= \sum_{k=0}^n\sum_{l=0}^k\left(\frac{\partial F}{\partial y^{(k)}}\right)_{i-k}a_{kl}\Delta x^{-k} \\ &= \sum_{k=0}^n\sum_{l=0}^k\left(\frac{\partial F}{\partial y^{(k)}}\right)_{i-k}a_{kl}\Delta x^{-k} \\ &= \sum_{k=0}^n(-1)^k\left[\frac{d^k}{dx^k}\left(\frac{\partial F}{\partial y^{(k)}}\right)\right]_i \end{align} $$ where I identified the backward finite difference: $$ \left[\frac{d^k}{dx^k}\Phi\right]_i = \frac{1}{(-\Delta x)^k}\sum_{l=0}^ka_{kl}\Phi_{i-l} $$

This generalises to other finite difference schemes, in particular the central one where you get the same finite difference by symmetry. In general, the integration by parts translates into an Abel transform in the discrete case, which is why the order is reversed.

Hope this helps.

Answer 2

Consider the functional $$ J[y] = \int_{x_1}^{x_2} F(x, y, y', \ldots, y^{(n)}) \,dx, $$ where $y = y(x)$. what we want is $$ \frac{\partial F}{\partial y} +\frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) - \frac{d^2}{dx^2}\left(\frac{\partial F}{\partial y''}\right) + \ldots + (-1)^n \frac{d^n}{dx^n}\left(\frac{\partial F}{\partial y^{(n)}}\right) = 0. $$

Let $y + \epsilon \eta$ be a variation of the function $y(x)$, where $\epsilon$ is a small parameter, and $\eta$ is a smooth function with $\eta(x_1) = \eta(x_2) = 0$. and The powers of $\eta(x_1) ,\eta(x_2)$ will also be zero for powers less than $n-1$

The variation of the functional is given by $$ J[y + \epsilon \eta] = \int_{x_1}^{x_2} F(x, y + \epsilon \eta, (y + \epsilon \eta)', \ldots, (y + \epsilon \eta)^{(n)}) \,dx. $$

Now, we consider the Taylor expansion of $F$ up to the first-order term in $\epsilon$: $$ F(x, y + \epsilon \eta, \ldots) = F(x, y, \ldots) + \epsilon \eta \frac{\partial F}{\partial y} + \epsilon \eta' \frac{\partial F}{\partial y'} + \ldots $$

Substitute this expansion into the integral: \begin{align*} J[y + \epsilon \eta] &= \int_{x_1}^{x_2} \left[F(x, y, \ldots) + \epsilon \eta \frac{\partial F}{\partial y} + \epsilon \eta' \frac{\partial F}{\partial y'} + \ldots \right] \,dx \\ &= J[y] + \epsilon \int_{x_1}^{x_2} \eta \frac{\partial F}{\partial y} \,dx + \epsilon \int_{x_1}^{x_2} \eta' \frac{\partial F}{\partial y'} \,dx + \ldots \end{align*}

Now, we apply the fundamental lemma of the calculus of variations, which states that if a smooth function $\eta$ satisfies $\int_{x_1}^{x_2} \eta \psi \,dx = 0$ for all smooth functions $\psi$ with $\psi(x_1) = \psi(x_2) = 0$, then $\eta$ must be identically zero.

By applying this lemma to each term in the expansion, we obtain conditions for the variation to vanish: \begin{align*} \int_{x_1}^{x_2} \eta \frac{\partial F}{\partial y} \,dx &= 0, \\ \int_{x_1}^{x_2} \eta' \frac{\partial F}{\partial y'} \,dx &= 0, \\ & \vdots \end{align*}

These conditions lead to the Euler-Lagrange equation: $$ \frac{\partial F}{\partial y} +\frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) - \frac{d^2}{dx^2}\left(\frac{\partial F}{\partial y''}\right) + \ldots + (-1)^n \frac{d^n}{dx^n}\left(\frac{\partial F}{\partial y^{(n)}}\right) = 0. $$

This equation characterizes the extremal functions for the given functional $J[y]$.

I Hope it Helps , Its easy to follow as well as Intutive